= Solution
{c}
<Eisenstein's criterion> says that an integer polynomial with leading coefficient not divisible by a prime $\ell$, all other coefficients divisible by $\ell$, and constant coefficient not divisible by $\ell^2$ is irreducible over the rationals, provided it is primitive; the usual <Gauss's lemma> supplies the passage from integer to rational factors.
Apply it with $\ell=2$ to
$$
(x+1)^8+1=x^8+8x^7+28x^6+56x^5+70x^4+56x^3+28x^2+8x+2.
$$
All nonleading coefficients are even and the constant is not divisible by four. A translation of the variable preserves reducibility, so \b[$x^8+1$ is irreducible over $\mathbb Q$].
Let $\zeta=e^{i\pi/8}$ and $c=\cos(\pi/8)$. The minimal polynomial of $\zeta$ is $x^8+1$, so $[\mathbb Q(\zeta):\mathbb Q]=8$. Since $2c=\zeta+\zeta^{-1}$, the number $\zeta$ satisfies $t^2-2ct+1=0$ over $\mathbb Q(c)$. This real field cannot contain the nonreal $\zeta$, so that extension has degree exactly two. The <tower law for field extensions> gives
$$
\boxed{[\mathbb Q(\zeta):\mathbb Q(c)]=2,\qquad
[\mathbb Q(c):\mathbb Q]=4.}
$$
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